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ENCE 4610Foundation Analysis and

Design

Shallow Foundations: Part I

Topics for Shallow Foundations, Part I

● Types of Shallow Foundations

● Spread Footing Design Concept and Procedure

● Bearing Capacity Failure Mechanisms Bearing Capacity

Equation Formulation Bearing Capacity

Correction Factors

● Other Items Local or Punching Shear Factors of Safety Practical Aspects of

Bearing Capacity Formulations

Presumptive Bearing Capacities

Types of Shallow Foundations

● Shallow foundations are usually placed within a depth D beneath the ground surface less than the minimum width B of the foundation

● Shallow foundations consist of:– Spread and continuous footings– Square, Rectangular or Circular

Footings– Continuous footings– Ring Foundations– Strap Footings

● Wall footings● Mats or Rafts

Footings

A finite spread footing is a shallow foundation that transmits loads and has an aspect ratio of 1 < L/B < 10

A continuous spread footing is an “infinite” footing where L/B > 10 and the effects of L are ignored

Abutment/Wingwall Footing

● A situation where the shallow foundation is a part of and acts with a retaining wall for both vertical load-bearing and horizontal loads of retained soil

Combined Footing● Combined footings are similar to

isolated spread footings except that they support two or more columns and are rectangular or trapezoidal in shape (Figure 8-7). They are used primarily when the column spacing is non-uniform (Bowles, 1996) or when isolated spread footings become so closely spaced that a combination footing is simpler to form and construct. In the case of bridge abutments, an example of a combined footing is the so-called “spill-through” type abutment (Figure 8-8). This configuration was used during some of the initial construction of the Interstate Highway System on new alignments where spread footings could be founded on competent native soils. Spill-through abutments are also used at stream crossings to make sure that foundations are below the scour depth of the stream.

Mat FoundationsMat Foundations A mat is continuous in two

directions capable of supporting multiple columns, wall or floor loads. It has dimensions from 20 to 80 ft or more for houses and hundreds of feet for large structures such as multi-story hospitals and some warehouses

Ribbed mats, consisting of stiffening beams placed below a flat slab are useful in unstable soils such as expansive, collapsible or soft materials where differential movements can be significant (exceeding 0.5 inch).

Bearing Pressure DistributionBearing Pressure Distribution

Concentric LoadsConcentric Loads

Flexible foundation on clay

Flexible Foundation on Sand

Rigid foundation on clay

Rigid Foundation on Sand

Simplified Distribution

Shear Failure vs. Settlement in Allowable Bearing Capacity

Plasticity: Lower and Upper Bound Solutions

● The Problem– Bearing Capacity failure is a plastic failure of the soil

along some failure surface– The problem of plastic failure of twofold:

● Finding the failure surface along which the plasticity an ultimately failure takes place

● Determining the failure state to which we should design, i.e. lower or upper bound

● The first is done by determining which failure surface provides the “path of least resistance” of failure

● The second is in part driven by uncertainty requirements in failure

● Review of Upper and Lower Bound Concept– Lower Bound: The true failure load is larger than the

load corresponding to an equilibrium system. The system has failed in at least one place.

– Upper Bound: The true failure load is smaller than the load corresponding to a mechanism, if that load is determined using the virtual work principle. The system has failed “in general.”

– The idea is that the true solution is somewhere between the two

● We saw this when we went through unsupported cuts in purely cohesive soils– In that case, we had to consider both the shape of

the failure surface and its location– For slopes, a circular failure surface was

considered as the most likely failure surface– The actual surface could be located for simple

slopes using theoretical considerations, but for more complex slopes (layered soils, water table, frictional soils) a trial and error solution was adopted

● In principle, only applicable to purely cohesive soils without friction, due to volume expansion considerations– Upper and lower bound theory can be extended to

soils with a frictional component (or only a frictional component,) but the implementation is much more complicated

● We will begin by considering strip (infinite or continuous) foundations only) in cohesive soils

Lower Bound Solution

By direct application and Mohr's Circle

By theory of elasticity

The more realistic lower bound

Upper Bound Limit Equilibrium MethodUpper Bound Limit Equilibrium Method

(Circular Failure Surface, Cohesive Soil)(Circular Failure Surface, Cohesive Soil)0

22 0 =B

BbσBcBbπcBbB

Bbq=M ultp

Assume: No soil strength due to internal friction (cohesive soil,) shear strength above foundation base neglectedWe add the effect of the weight of the soil (effective stress) acting on the top of the right

side of the circle against rotation.

0cult

c

0ult

σ+cN=q

πcBb=N

σ+cπcBb=q

6.282

2

More Realistic Upper Bound Case

● This is done by moving the centre of the failure surface upward

● A circular failure surface is what we assumed for slope failure

● It is valid for very soft clays, and methods have been developed for use with these types of soils

● Soft clays are more subject to settlement

● We will not use these in this course

So the solution is bounded by

Development of Prandtl Bearing Capacity Theory

● Application of limit equilibrium methods first done by Prandtl on the punching of thick masses of metal (materials with no internal frictional effects)

● Prandtl's methods first adapted by Terzaghi to bearing capacity failure of shallow foundations (specifically, he added the effects of frictional materials)

● Vesić and others (Meyerhof, Brinch Hansen, etc.) improved on Terzaghi's original theory and added other factors for a more complete analysis

Note the three zones, the foundation fails along the lower boundary of these zones

Assumptions for Bearing Assumptions for Bearing Capacity MethodsCapacity Methods

Geometric assumption– Depth of foundation is less than

or equal to its width– Foundation is a strip footing

(infinite length)*

Geotechnical Assumptions– Soil beneath foundation is

homogeneous semi-infinite mass*– Mohr-Coulomb model for soil– General shear failure mode is the

governing mode*– No soil consolidation occurs– Soil above bottom of foundation

has no shear strength; is only a surcharge load against the overturning load*

Foundation-Soil Interface Assumptions Foundation is very rigid

relative to the soil No sliding occurs

between foundation and soil (rough foundation)

Loading Assumptions Applied load is

compressive and applied vertically to the centroid of the foundation*

No applied moments present*

* We will discuss “workarounds” to these assumptions

Loads and Failure Zones Loads and Failure Zones on Strip Foundationson Strip Foundations

Q = load/unit length on foundationq = load/unit area due to effective stress at base of foundationI, II, III = regions of failure in Prantdl theory for bearing capacity failure

Basic Equation of Bearing Capacity

Basic Equation of Bearing Capacity

● Values of Nc, Nq mostly the same. Values of Nγ depend upon theory

● DIN/Brinch-Hansen:

● CFEM:

● Vesić:

• Vesić-AASHTO Factors:

Bearing Capacity Example

Types of Bearing Capacity Traditionally, bearing capacity has

been classified as follows: General Shear (the case

upon which the theory is based)

Local Shear Punching Shear Which one takes place

depends upon consistency or density of soil, which decreases from general to local to punching

Generally, with softer soils, settlement tends to govern the design more than bearing capacity

Soil Property Corrections for Local and Punching Shear

● Local and punching shear are accounted for by reducing the cohesion and/or friction angle of the soil

● Read entire section on how and when to use this reduction

● Also analyze for settlement

Bearing Capacity Correction Factors

Shape FactorsShape Factors

Inclined Base Factors

Groundwater Table Correction Factors

Embedment Depth Factors

Load Inclination Factors A convenient way to account for the

effects of an inclined load applied to the footing by the column or wall stem is to consider the effects of the axial and shear components of the inclined load individually. If the vertical component is checked against the available bearing capacity and the shear component is checked against the available sliding resistance, the inclusion of load inclination factors in the bearing capacity equation can generally be omitted. The bearing capacity should, however, be evaluated by using effective footing dimensions, as discussed in Section 8.4.3.1 and in the footnote to Table 8-4, since large moments can frequently be transmitted to bridge foundations by the columns or pier walls. The simultaneous application of shape and load inclination factors can result in an overly conservative design.

Unusual column geometry or loading configurations should be evaluated on a case-by-case basis relative to the foregoing recommendation before the load inclination factors are omitted. An example might be a column that is not aligned normal to the footing bearing surface. In this case, an inclined footing may be considered to offset the effects of the inclined load by providing improved bearing efficiency (see Section 8.4.3.4). Keep in mind that bearing surfaces that are not level may be difficult to construct and inspect. (FHWA NHI-06-089)

Allowable Bearing Capacity

Bearing Capacity Example

Given Square shallow foundation,

5’ x 5’ Foundation depth = 2’ Cohesionless Unit weight 121 pcf Internal friction angle 31

degrees Load on foundation = 76

kips Groundwater table very deep

Find Factor of safety against

bearing capacity failure Solution

Governing equation:

We can neglect factors due to groundwater table (Cw), load inclination (b) and depth (d)

Bearing Capacity Example

Bearing capacity “N” factors for 31 degree friction angle Nc = 32.7 Nq = 20.6 Nγ = 26.0

Bearing Capacity Example

● Shape Factors● sc = 1+(5/5)(20.6/32.7)

= 1.63● sγ=1-0.4(5/5) = 0.6● sq = 1+(5/5)(tan(31)) =

1.6

Bearing Capacity Example

Other variables c=0 (problem statement) q=(121)(2) = 242 psf γ= 121 pcf (problem

statement) Substitute and solve

qult = (0) + (242)(20.6)(1.6) + (0.5)(121)(5)(26.0)(0.6) = 0 + 7976.32 + 4719 = 12,695 psf

Compute ultimate load Qult = qult * A (12695)(5)(5) = 317,383

lbs. = 317.3 kips Compute Factor of Safety

FS = 317.3/76 = 4.17 It’s also possible to do this

using the pressures qa = 76,000/(25) = 3040

psf FS = 12,695/3040 = 4.17

Effect of Groundwater Table and Layered Soils on Bearing Capacity

Layered Soils are virtually unavoidable in real geotechnical situations

Softer layers below the surface can and do significantly affect both the bearing capacity and settlement of foundations

Pore water pressure increases; reduces both effective stress and shear strength in the soil (same problem as is experienced with unsupported slopes)

● Three ways to analyze layered soil profiles:

Use the lowest of values of shear strength, friction angle and unit weight below the foundation. Simplest but most conservative. Use groundwater factors in conjunction with this.

Use weighted average of these parameters based on relative thicknesses below the foundation. Best balance of conservatism and computational effort. Use width of foundation B as depth for weighted average

Consider series of trial surfaces beneath the footing and evaluate the stresses on each surface (similar to slope failure analysis.) Most accurate but calculations are tedious; use only when quality of soil data justify the effort

● Groundwater considered using the groundwater correction factor

Groundwater Example

● Given Previous example Groundwater table is 3’ below

base of foundation● Find

Bearing Capacity● Solution

Note that groundwater factor Cwγ is based on Dw, which is distance from surface of soil to groundwater table

Dw = 2’ + 3’ = 5’ Foundation depth plus

distance below the base of the foundation

● Solution Values for interpolation (from Table 8-

5) Dw = Df = 2’ (base of foundation):

Cwq = 1.0 Cwγ = 0.5

Dw = 1.5 * Bf + Df = (1.5)(5’) + 2’ =9.5’ (bottom of influence zone): = 1.0 Cwq = 1.0 Cwγ = 1.0

Interpolating: Cwq = 1.0 Cwγ = 0.7

● Substitute and solve qult = (0) + (242)(20.6)(1.6)(1.0) + (0.5)

(121)(5)(26.0)(0.6)(0.7) = 0 + 7976.32 + 3303.3 = 11,280 psf (11% decrease)

The Other “Workarounds” for Bearing Capacity Theory

● We have already discussed workarounds to the following: Strip footing (shape factors) Level base (base inclination

factors) Vertical load (load

inclination factors) Homogeneous soil

(groundwater factors and related theory)

General Shear (modified c and φ values)

● Other workarounds we will discuss are as follows: Load eccentricity (load is off

centre or there is a moment that accompanies the load Load eccentricity is

unavoidable in some circumstances because of the geometry of the structure and the site

Eccentricity not only impacts bearing capacity but also the basic stability of the foundation base (foundation liftoff)

Footings on top of or on slopes

Eccentric Loading of Foundations● Eccentric loading occurs when

a footing is subjected to eccentric vertical loads, a combination of vertical loads and moments, or moments induced by shear loads transferred to the footing.

Abutments and retaining wall footings are examples of footings subjected to this type of loading condition.

Moments can also be applied to interior column footings due to skewed superstructures, impact loads from vessels or ice, seismic loads, or loading in any sort of continuous frame.

● Eccentricity is the distance from the effective point of loading to the centroid of the foundation.

This distance can be one-way (strip and circular footings) or two-way (square or rectangular footings)

Eccentricity can occur either because the loading is not at the centroid or there is a moment on the foundation.

In the case of a moment, the eccentricity is computed by dividing the moment on the foundation by the applied load, i.e., e = M/P

Ways of Accounting for Eccentricity

Expressing Load Eccentricity and Inclination

Load Divided by Inclination Angle

Total Load P Total Vertical Load Pv Total Horizontal Load Ph Angle of Inclination α =

arctan(Ph/Pv) Load can be concentric or

eccentric Load and Eccentricity

Total Vertical Load P Eccentricity from centroid

of foundation e Horizontal Load (if any)

not included

Moment and Eccentric Load

Total eccentric vertical Load P with eccentricity e

Replace with concentric vertical load P and eccentric moment M=Pe

Continuous Foundations Moments, loads

expressed as per unit length of foundation, thus P/b or M/b

Eccentricity and

Equivalent Footing

Procedure

One Way Loading

One-way loading is loading along one of the centre axes of the foundation

Three cases to consider (see right)

Resultant loads outside the “middle third” result in foundation lift-off and are thus not permitted at all

After this reduced footing size can be computed

Example of One-Way Eccentricity● Given

Continuous Foundation as shown

Groundwater table at great depth

Weight of foundation (concrete) not included in load shown

● Find Whether resultant force

acts in middle third Minimum and maximum

bearing pressures Reduced bearing area

Example of One Way Example of One Way EccentricityEccentricity

Compute Weight of Foundation Wf/b = (5)(1.5)(150) = 1125 lb/ft

Compute eccentricity

Thus, eccentricity is within the “middle third” of the foundation and foundation can be analysed further without enlargement at this point

e=(M /b )

Q /b=

800012000+1125

=0 .61 ft .

B6=

56=0 .833 ft .>0 .61 ft .

Equations for One-Way Pressures Equations for One-Way Pressures

with Eccentric/Moment Loadswith Eccentric/Moment Loads

If q at any point is less than zero, resultant is outside the middle third

Wf is foundation weight

Example of One Way Example of One Way EccentricityEccentricity

Compute minimum and maximum bearing pressures

Example of One-Way Eccentricity

● Since the resultant is within the middle third, we can compute the reduced foundation size B’

● As this is a continuous footing experiencing one-way eccentricity, we do not need to consider an L’

● From previous computations: B = 5’ e = 0.61’

● Reduced Foundation Width B’ = 5 – (2)(0.61) = 3.78’

Two-Way Eccentricity• Eccentricity in both “B” and “L”

directions produces a planar distribution of stress

• Kern of Stability Foundation stable

against overturn only if resultant falls in the kern in the centre of the foundation

Resultant in the kern if

6 eBB

+6 eLL

≤1

eB, e

L = eccentricity in B, L directions

Bearing Pressure at CornersBearing Pressure at CornersTwo-Way EccentricityTwo-Way Eccentricity

• Helpful hint to prevent confusion of eccentricity of finite vs. infinite (continuous) foundationso Always use one-way

eccentricity equations for continuous foundations

o Always use two-way eccentricity equations for finite foundations

o Two-way equations will reduce to one-way equations if one of the eccentricities (eB, eL) is zero

L

e

B

e

BL

WQq LBf 66

14,3,2,1

Two-Way Eccentricity Example Given

Grain silo design as shown Each silo has an empty

weight of 29 MN; can hold up to 110 MN of grain

Weight of mat = 60 MN Silos can be loaded

independently of each other Find

Whether or not eccentricity will be met with the various loading conditions possible

Eccentricity can be one-way or two-way

Two-Way Eccentricity Two-Way Eccentricity ExampleExample

One-Way Eccentricity Largest Loading: two

adjacent silos full and the rest empty

Q = (4)(29) + 2(110) + 60 = 396 MN

M = (2)(110)(12) = 2640 MN-m

e=MQ

e=2640396

e=6 .67mB6

=506

=8 .33m>6. 67m

Eccentricity OK for one-way eccentricity

Two-Way Eccentricity Two-Way Eccentricity ExampleExample

Two-Way Eccentricity Largest Loading: one silo full and the rest

empty P = (4)(29) + 110 + 60 = 286 MN MB = ML = (110)(12) = 1320 MN-m

eB=eL=MQ

=1320286

=4 .62m

6 eBB

+6eLL

=2( ( 6 ) ( 4 .62 )

50 )=1 .11>1

Not acceptable

Two-Way Eccentricity Two-Way Eccentricity ExampleExample

Two-Way Eccentricity Solution to Eccentricity

Problem: increase the size of the mat

Necessary to also take other considerations into account (bearing failure, settlement, etc.)

6 eBB

+6eLL

=2( ( 6 ) ( 4 .62 )

B )=1

B=L=55. 4m

Equivalent Footing Using Two-Equivalent Footing Using Two-

Way Eccentricity ExampleWay Eccentricity Example Largest Loading: one silo full and the rest

empty

Result of Two-Way Eccentricity Analysis

eB = eL = 4.62 m B = L = 55.4 m (expanded

foundation) Equivalent Footing

Dimensions B’ = B – 2eB = 55.4 – (2)(4.62) B’ = 45.8 m = L’ (as B = L and

eB = eL)

Equivalent Footing Using Two-Equivalent Footing Using Two-

Way Eccentricity ExampleWay Eccentricity Example

One-Way Eccentricity Largest Loading: two

adjacent silos full and the rest empty

B = L = 55.4 m (expanded foundation)

eB = 6.67m eL = 0 m B’ = B’-2eB = 55.4 – (2)

(6.67) = 42.1 m L = L’ = 55.4 m B

6=

506

=8 .33m>6. 67m

Other Notes on Bearing Other Notes on Bearing Capacity FactorsCapacity Factors

• Two ways to handle B’ and L’ values when computing shape factors (which are a function of B/L):o AASHTO (2002) guidelines recommend calculating the shape factors, s, by using the effective

footing dimensions, B and L’.′ and L’.o However, the original references (e.g., Vesi , 1975) do not specifically recommend using the ć, 1975) do not specifically recommend using the

effective dimensions to calculate the shape factors. Since the geotechnical engineer typically does not have knowledge of the loads causing eccentricity, use full footing dimensions be used to calculate the shape factors for use in computation of ultimate bearing capacity.

o Either is acceptable for problems in this class. In practice, which one you would use would depend upon a) the project (a highway project would tend to use AASHTO recommendations) and how well the location of the loads was known.

• Bowles (1996) also recommends that the shape and load inclination factors (s and i) should not be combined.

• In certain loading configurations, the designer should be careful in using inclination factors together with shape factors that have been adjusted for eccentricity (Perloff and Baron, 1976). The effect of the inclined loads may already be reflected in the computation of the eccentricity. Thus an overly conservative design may result.

Bearing Capacity for Foundation at Top of a SlopeBearing Capacity for Foundation at Top of a Slope

Example of Footings Example of Footings on Slopeson Slopes

• Solutiono Ncq for D/B = 0 and Slope Angle

of 30 deg. = 4.9

o Ncq for D/B = 1 and Slope Angle of 30 deg. = 6.4

o Linearly interpolating, Ncq =

(6.4+4.9)/2 = 5.7

o Nγq = 1 since the soil is purely cohesive

o B/2 = Do Shape factors are unity because it

is a continuous footing; water table and embedment factors are like wise not considered

– qult = (75)(5.7)+(1.5)(18.5) =

455.25 kPa

• Giveno Strip footing to be constructed on top

of the slopeo Soil properties: c = 75 kPa, γ = 18.5

kN/m3, water table very deepo H = 8 m, B = 3 m, D = 1.5 m, b = 2

m, Slope Angle = 30 deg.

• Findo Ultimate Bearing Capacity of

Footing, using solution method of previous slide

o Solutiono B < H since 3 m < 8 mo Obtain Ncq from Figure 8-18(e) for

Case I with No = 0o D/B = 1.5/3 = 0.5o b/B = 2/3 = 0.667

Required Footing Required Footing SetbacksSetbacks

Practical Aspects of Bearing Capacity Formulations

Failure Zones for Bearing Capacity

Presumptive Bearing CapacityPresumptive Bearing Capacity

Presumptive Bearing Capacity on Rock

Questions?